Question

Use implicit differentiation to find an equation of the tangent line to the curve, called a devil's curve, at the point (0, -2).Use implicit differentiation to find an equation of the tangent line to the curve, called a devil's curve, at the point (0, -2).y2 (y2-4) = x2 (x2 - 5)

Answer 1

The equation of the tangent line to the curve is y = -2.

Given,

         $y^{2} (y^{2} -4)= x^{2} (x^{2} -5)$

On solving the above equation, we get

          $y^{4} - 4y^{2}=x^{4}-5x^{2}$

Differentiate with respect to x.

       $4y^{3} y^{'} -8yy^{'}=4x^{3}-10x$

       $y^{'}(4y^{3}-8y )=4x^{3} -10x$

                      $y^{'} = \frac{4x^{3}-10x }{4y^{3}-8y }$

To find the slope of the tangent line at (0,-2) plug x = 0 and y = -2 into the $y^{'}$.

                     $y^{'} = \frac{0}{-16}$

Hence,

         The slope is 0 and therefore a horizontal line with a y value of -2, i.e. y = -2.

The equation of the tangent line to the curve is y = -2.

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